Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive.  What is the probability that they each have a different tens digit?
Answer: In this set of integers, there are 5 tens digits: {2, 3, 4, 5, 6}.  If 5 integers all have different tens digits, then there must be exactly one integer among the 5 with each tens digit.  Since there are 10 different integers for each tens digit, the number of ways to pick, without regard to order, 5 different integers with different tens digits is $10^5$.  The total number of combinations of 5 integers is $\binom{50}{5}$.  So the probability that 5 integers drawn all have the different tens digits is $$ \frac{10^5}{\binom{50}{5}} = \frac{100000}{2118760} = \boxed{\frac{2500}{52969}}. $$